GALOIS POINTS ON VARIETIES
MOSHE JARDEN AND BJORN POONEN
Abstract. A field K is ample if for every geometrically integral K-variety V with a smooth
K-point, V (K) is Zariski dense in V . A field K is Galois-potent if every geometrically
integral K-variety has a closed point whose residue field is Galois over K. We prove that
every ample field is Galois-potent. But we construct also non-ample Galois-potent fields; in
fact, every field has a regular extension with these properties.
1. Introduction
Definition 1.1. Let X be a variety over a field K. By a Galois point on X, we mean a
closed point whose residue field is Galois over K. We say that K is Galois-potent if every
geometrically integral K-variety has a Galois point.
Question 1.2. Is every field Galois-potent?
We do not even know if Q is Galois-potent. On the other hand, we have the following
definition of Florian Pop:
Definition 1.3 (cf. [Pop96, p. 2] and [Jar11, Chapter 5]). A field K is called ample (or large
or anti-Mordellic) if for every geometrically integral K-variety V with a smooth K-point,
V (K) is Zariski dense in V .
Pseudo-algebraically closed (PAC) fields and Henselian fields (e.g., Qp ) are ample; see
[Jar11, Chapter 5] for these and many more examples. Our first main theorem is the following:
Theorem 1.4. Ample fields are Galois-potent.
Some non-ample fields too are Galois-potent. For example, finite fields are non-ample, but
have abelian absolute Galois group and hence trivially are Galois-potent. Less trivially, we
can also construct infinite non-ample fields that are Galois-potent: in Section 5 we will prove
the following.
Theorem 1.5. Every field admits a regular extension that is Galois-potent but not ample.
Bary-Soroker and Fehm in [BSF13, Section 2.2] write that “all infinite non-ample fields
appearing in the literature are Hilbertian”. The non-ample fields we construct in the proof
of Theorem 1.5 turn out to be Hilbertian too (Remark 5.4), and in particular they have
non-abelian absolute Galois group.
Date: November 26, 2015.
2010 Mathematics Subject Classification. Primary 12E30; Secondary 11G35, 14G99.
The first author was supported by the Minkowski Center for Geometry at Tel Aviv University, established
by the Minerva Foundation. The second author was supported in part by National Science Foundation grant
DMS-1069236 and a grant from the Simons Foundation (#340694 to Bjorn Poonen). Any opinions, findings,
and conclusions or recommendations expressed in this material are those of the authors and do not necessarily
reflect the views of the funding agencies.
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Question 1.6. Are there also infinite non-ample fields with abelian absolute Galois group?
If the answer to Question 1.6 were yes, it would immediately yield another example of
a non-ample Galois-potent field. But there are several questions and conjectures in the
literature that each imply a negative answer to Question 1.6, as we now explain. Let GK be
the absolute Galois group of K.
• Is every infinite non-ample field Hilbertian? (Cf. [BSF13, Section 2.2] again.) A Hilbertian
field K cannot have an abelian GK .
• Is every infinite field K with topologically finitely generated GK ample? (See the paragraph
before Theorem 5.5 of [JK10], or see [BSF13, 4.2, Question III].) By [Koe01], if GK is
abelian, either K is henselian and hence ample, or GK is procyclic and in particular
topologically finitely generated, and hence ample if the question at the start of this bulleted
item has a positive answer.
• Is every infinite field with prosolvable GK ample? (See [BSF13, 4.1, Question II].) If so,
then every infinite field with abelian GK is ample too.
2. Notation
Let K be a field. Let K be an algebraic closure of K. By a K-variety we mean a separated
scheme of finite type over K. Given a K-variety X and a finite extension L ⊃ K, let XL be
the L-variety X ×K L; similarly, if φ is a K-morphism, let φL be its base change. If X is
an integral K-variety, let K(X) denote its function field. By a K-curve, we mean a smooth
geometrically integral K-variety of dimension 1. By the absolute genus gX of a K-curve X, we
mean the genus of the smooth projective model of XK . Given a curve X and n ≥ 1, let X (n)
denote the nth symmetric power, defined as the (variety) quotient of X n by the symmetric
group Sn .
3. Restriction of scalars
Let K ⊆ L be a finite extension of fields. For any quasi-projective L-variety X, the functor
sending each K-scheme S to the set X(SL ) is representable by a K-variety ResL/K X called
the restriction of scalars (cf. [BLR90, §7.6, Theorem 4]). If, in addition, L is separable over
K, then ResL/K X after base field extension becomes isomorphic to a product of [L : K]
conjugates of X; in particular, if X is geometrically integral and smooth of dimension d over
L, then ResL/K X is geometrically integral and smooth of dimension [L : K]d over K.
4. Ample fields are Galois-potent
In this section we prove Theorem 1.4.
Proposition 4.1. Let K be an ample field. Let X be a geometrically integral K-variety. Let
N be a finite separable extension of K. If X has a smooth N -point, then X has a point with
residue field N .
Proof. Replacing X by a Zariski-open subvariety, we may assume that X is smooth and
affine. For each finite separable extension L of K, define X L := ResL/K (XL ). Thus X L (S) =
XL (SL ) = X(SL ) for any K-scheme S, and in particular, X L (K) = X(L). If L ⊆ L0 are two
0
such extensions, the natural map X(SL ) → X(SL0 ) can be rewritten as X L (S) → X L (S),
0
and it is functorial in S, so Yoneda’s lemma yields a K-morphism X L → X L .
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We have X N (K) = X(N ), which is nonempty by assumption. Also, K is ample, and X N is
smooth and geometrically integral, so X N (K) is Zariski-dense in X N . There are only finitely
many fields L with K ⊆ L ( N , and for each such L, we have dim X L < dim X N . Thus X N
has a K-point outside the image of every morphism X L → X N . In other words, X(N ) has a
element outside X(L) for every L. For this element, the image of Spec N → X is a closed
point with residue field N .
Proof of Theorem 1.4. Let K be an ample field, and let X be any geometrically integral
K-variety. Choose a finite separable extension N of K such that X(N ) 6= ∅. Enlarge N to
assume that N is Galois over K. By Proposition 4.1, X has a closed point with residue field
N.
5. Infinite Galois-potent fields that are not ample
In this section we prove Theorem 1.5.
Lemma 5.1. Let K be an algebraically closed field. Let X, Y, C be K-curves with gC > 1.
Any rational map X × Y 99K C factors through one of the projections to X or Y .
Proof. A rational map φ : X × Y 99K C may be viewed as an algebraic family of rational
maps X 99K C parametrized by (an open subvariety of) Y . But the de Franchis–Severi
theorem [Sam66, Théorème 1] implies that there are no nonconstant algebraic families of
nonconstant rational maps X 99K C. Thus either the rational maps in the family are all the
same, in which case φ factors through the first projection, or each rational map in the family
is constant, in which case φ factors through the second projection.
Lemma 5.2. Let X be a curve over a field K. Let F = K(X (2) ). Then X has a point over
a quadratic extension of F , and C(F ) = C(K) for every K-curve C of absolute genus > 1.
Proof. Either projection X × X → X gives a point of X over the quadratic extension
K(X × X) of K(X (2) ) = F .
Let c ∈ C(F ). Then c corresponds to a rational map X (2) 99K C. Composing X ×X → X (2)
with such a rational map yields a rational map φ : X × X 99K C. By Lemma 5.1, φK is
constant on all vertical copies of XK or constant on all horizontal copies of XK . Since φK is
S2 -invariant, it is constant on all vertical and horizontal copies of XK . Thus φK is constant.
Hence φ is constant. Equivalently, c ∈ C(K).
Lemma 5.3. Every field K admits a regular extension K 0 such that
(i) Every K-curve X has a point over an at most quadratic extension of K 0 (possibly
depending on X).
(ii) For every K-curve C of absolute genus > 1, we have C(K 0 ) = C(K).
Proof. Let (Xα )α<τ be a well-ordering of the set of K-curves up to isomorphism, indexed by
an ordinal τ . For α ≤ τ , define Kα by transfinite induction as follows:
• Let K0 := K.
(2)
• For each α < τ , define Kα+1 := Kα (Xα );
• If α is a limit ordinal, define Kα := limβ<α Kβ .
−→
Let K 0 := Kτ . Then K 0 is regular over K by [FJ08, Corollary 2.6.5(d)], whose proof remains
valid when the cardinal m is replaced by an ordinal τ .
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(i) Any K-curve X is Xα for some α < τ . Lemma 5.2 shows that X has a point over a
quadratic extension of Kα+1 , and hence also a point over an at most quadratic extension
of K 0 .
(ii) Let C be a K-curve of absolute genus > 1. By Lemma 5.2 and induction, C(Kα ) = C(K)
for all α ≤ τ . In particular, C(K 0 ) = C(K).
Proof of Theorem 1.5. Let K be the given field. Construct a sequence of fields L0 , L1 , . . .
inductively as follows: let L0 := K(t) with t transcendental over K, and let Li+1 be the
regular extension of Li given by Lemma 5.3. Let L∞ := lim Li .
−→
Let X be an L∞ -curve. Then X is definable over some Li . The conditions in Lemma 5.3
imply that X has a point over an at most quadratic extension of Li+1 , and hence a point
over an at most quadratic extension of L∞ . Every geometrically integral L∞ -variety other
than a point contains an L∞ -curve, so L∞ is Galois-potent.
Choose a non-isotrivial curve C of absolute genus > 1 over L0 = K(t) such that C has a
smooth L0 -point; then C(L0 ) is finite [Sam66, Théorème 4]. The conditions in Lemma 5.3
imply that C(L0 ) = C(L1 ) = · · ·, so C(L∞ ) = C(L0 ), which is finite. Thus L∞ is not
ample.
Remark 5.4. The field L∞ constructed in the proof of Theorem 1.5 is Hilbertian, because of
the following two facts:
(a) Any finitely generated transcendental extension of a field is Hilbertian [FJ08, Theorem 13.4.2].
(b) If (Kα )α≤τ is an ascending tower of fields indexed by an ordinal τ , and Kα is Hilbertian for
each α < τ , and Kα+1 is a regular extension of Kα for each α < τ , and Kα = limβ<α Kβ
−→
for each limit ordinal α ≤ τ , then Kτ is Hilbertian.
(Fact (b) appears as [FJ08, Chapter 12, Exercise 2], but the hypothesis Kα = limβ<α Kβ is
−→
missing there.)
Proof of (b). Given irreducible polynomials fi (~t, x) over Kτ for i = 1, . . . , r, each separable in
x, we must find infinitely many ~a over Kτ such that the specializations fi (~a, x) for i = 1, . . . , r
are irreducible over Kτ . There exists α < τ such that all the fi are defined over Kα . Since Kα
is Hilbertian, there exist infinitely many ~a over Kα such that fi (~a, x) for each i is irreducible
over Kα . These specializations are irreducible over Kτ as well, since Kτ is a regular extension
of Kα by [FJ08, Corollary 2.6.5(d)].
Remark 5.5. Padmavathi Srinivasan has constructed a field K with stronger properties than
the L∞ in the proof of Theorem 1.5, namely that C(K) is finite for every K-curve C of
absolute genus > 1, while there is a degree 2 field extension L ⊃ K that is ample [Sri15].
Acknowledgements
We thank the referee for many insightful comments and corrections.
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School of Mathematics, Tel Aviv University, Ramat Aviv, Tel Aviv 6139001, Israel
E-mail address: jarden@post.tau.ac.il
Department of Mathematics, Massachusetts Institute of Technology, Cambridge, MA
02139-4307, USA
E-mail address: poonen@math.mit.edu
URL: http://math.mit.edu/~poonen/
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